I can see that by applying the product rule, all resulting expressions containing $(x-6)$ would become zero, hence leaving $(1)(x-1)(x-2)(x-3)(x-4)(x-5)(x-7)(x-8)(x-9)(x-10)$, which, at $x=6$, is $2880$. However, I'm not sure how to formally write out a concise method for finding this solution.
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$\begingroup$ some summands of your derivative are zero if you plug in $$x=6$$ $\endgroup$– Dr. Sonnhard GraubnerCommented Oct 6, 2017 at 18:16
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1$\begingroup$ Your approach is correct. Your answer seems correct (it is wrong only if you have calculated it wrong). What is your question exactly? $\endgroup$– Aniruddha DeshmukhCommented Oct 6, 2017 at 18:17
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$\begingroup$ Is it incorrect to write out a solution in words only, without showing the mathematical analysis? If so, I'd like to be able to show mathematical analysis. $\endgroup$– DataCommented Oct 6, 2017 at 18:21
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2 Answers
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$P(x)=\prod\limits_{k=1}^{10}(x-k)=(x-6)Q(x)$ where $Q(x)=\prod\limits_{k=1\\k\neq 6}^{10}(x-k)$
$P'(x)=Q(x)+(x-6)Q'(x)$
Thus $P'(6)=Q(6)+0=Q(6)=2880$
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2$\begingroup$ +1 This is easily the best answer. It may also help to note that $P'(6)=Q(6)+0=5!\cdot4!=2880$. I only mention this because I would hope that $5!\cdot4!$ would be an acceptable answer instead of actually computing $2880$ as an end result. $\endgroup$ Commented Oct 6, 2017 at 18:36
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$\begingroup$ @DanielW.Farlow yes, I did not detail this part since it did not seem to be an issue for the OP, but it is certainly the way to go to compute this value. $\endgroup$– zwimCommented Oct 6, 2017 at 19:06
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$\begingroup$ In fact, you don't even need $Q$ to be differentiable at $x=6$ to get the conclusion $P'(6) = Q(6)$ - with a direct calculation using the limit definition of derivative, you get it only assuming $Q$ is continuous at $x=6$. $\endgroup$ Commented Oct 6, 2017 at 21:13
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$\begingroup$ @DanielW.Farlow Why is $5!⋅4!$ preferable to $2880$? $\endgroup$– DataCommented Oct 7, 2017 at 11:05
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$\begingroup$ What a genius method!! $\endgroup$ Commented Sep 10, 2020 at 19:26
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Well, it depends on which properties you will take as given. Let's just say that from the usual product rule ($(fg)'=f'g+fg'$) you can obtain $$ (fgh)'=f'(gh)+f(gh)'=f'gh+fg'h+fgh',$$ and so on. In a general case, you could write $$\left(\prod_{k=1}^n f_k \right)'=\sum_{k=1}^n f_k' \cdot \prod_{j\not=k} f_j.$$
You may try from there (in your case, $f_k=x-k$ and n=10).
answered Oct 6, 2017 at 18:25
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$\begingroup$ Could you explain the index “$j\neq k$”? $\endgroup$ Commented Oct 9, 2017 at 23:41
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1$\begingroup$ Sure! I was actually being a little lazy. It should be $j$ from $1$ to $n$, and $j \neq k$. That is: you'll end up with an $n$-terms sum (because $k$ goes from $1$ to $n$), and in each of those terms you'll have a product of the derivative of $f_k$ and $n-1$ other factors, because you'll have $f_1 \cdot f_2 \cdot \ldots \cdot f_{k-1} \cdot f_{k+1} \cdot \ldots f_n$ ($f_1$ is omitted in the first term, $f_2$ in the second, and so on). Hope I didn't make it LESS clear. =S $\endgroup$ Commented Oct 10, 2017 at 0:45
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1$\begingroup$ Let's just say $n=3$, then this will become: $$(f_1 \cdot f_2 \cdot f_3)' = f'_1 \cdot f_2 \cdot f_3 + f'_2 \cdot f_1 \cdot f_3 + f'_3 \cdot f_1 \cdot f_2. $$ $\endgroup$ Commented Oct 10, 2017 at 0:46
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$\begingroup$ Oh, now I understand! I had mixed up the argument of $\sum$. It's just shorthand for $$\sum_{k\in S} \left( f_k^\prime \prod_{j\in S\setminus\{k\}}f_j \right)$$ where $$S=\bigcup_{i=1}^{n}\{i\}$$ $\endgroup$ Commented Oct 10, 2017 at 4:33